Numbers formed by flipping common set bits in two given integers

Given two positive integers A and B, the task is to flip the common set bitsin A and B.

Examples:

Input: A = 7, B = 4 
Output: 3 0 
Explanation:
The binary representation of 7 is 111 
The binary representation of 4 is 100 
Since the 3rd bit of both A and B is a set bit. Therefore, flipping the 3rd bit of A and B modifies A = 3 and B = 0 
Therefore, the required output is 3 0

Input: A = 10, B = 20 
Output: 10 20

Naive Approach: The simplest approach to solve this problem is to check if the i^th bit of both A and B is a set or not. If found to be true, then flip the i^th bit of both A and B. Finally, print the updated values of both A and B.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to flip bits of A and B
// which are set bits in A and B
void flipBitsOfAandB(int A, int B)
{
    // Iterator all possible bits
    // of A and B
    for (int i = 0; i < 32; i++) {

        // If ith bit is set in
        // both A and B
        if ((A & (1 << i)) && (B & (1 << i))) {

            // Clear i-th bit of A
            A = A ^ (1 << i);

            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }

    // Print A and B
    cout << A << " " << B;
}

// Driver Code
int main()
{
    int A = 7, B = 4;

    flipBitsOfAandB(A, B);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to flip bits of A and B
// which are set bits in A and B
static void flipBitsOfAandB(int A, int B)
{

    // Iterator all possible bits
    // of A and B
    for(int i = 0; i < 32; i++)
    {

        // If ith bit is set in
        // both A and B
        if (((A & (1 << i)) &
             (B & (1 << i))) != 0)
        {

            // Clear i-th bit of A
            A = A ^ (1 << i);

            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }

    // Print A and B
    System.out.print(A + " " + B);
}

// Driver Code
public static void main(String[] args)
{
    int A = 7, B = 4;

    flipBitsOfAandB(A, B);
}
}

// This code is contributed by code_hunt

Python3

# Python3 program to implement
# the above approach

# Function to flip bits of A and B
# which are set in both of them
def flipBitsOfAandB(A, B):

    # Iterate all possible bits of
    # A and B
    for i in range(0, 32):

        # If ith bit is set in
        # both A and B
        if ((A & (1 << i)) and
            (B & (1 << i))):

            # Clear i-th bit of A
            A = A ^ (1 << i)

            # Clear i-th bit of B
            B = B ^ (1 << i)

    print(A, B)

# Driver Code
if __name__ == "__main__" :

    A = 7
    B = 4

    flipBitsOfAandB(A, B)

# This code is contributed by Virusbuddah_

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to flip bits of A and B
// which are set bits in A and B
static void flipBitsOfAandB(int A, int B)
{

    // Iterator all possible bits
    // of A and B
    for(int i = 0; i < 32; i++)
    {

        // If ith bit is set in
        // both A and B
        if (((A & (1 << i)) &
             (B & (1 << i))) != 0)
        {

            // Clear i-th bit of A
            A = A ^ (1 << i);

            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }

    // Print A and B
    Console.Write(A + " " + B);
}

// Driver Code
public static void Main(string[] args)
{
    int A = 7, B = 4;

    flipBitsOfAandB(A, B);
}
}

// This code is contributed by chitranayal

Javascript

<script>
// javascript program to implement
// the above approach


// Function to flip bits of A and B
// which are set bits in A and B
function flipBitsOfAandB(A , B)
{

    // Iterator all possible bits
    // of A and B
    for(i = 0; i < 32; i++)
    {

        // If ith bit is set in
        // both A and B
        if (((A & (1 << i)) &
             (B & (1 << i))) != 0)
        {

            // Clear i-th bit of A
            A = A ^ (1 << i);

            // Clear i-th bit of B
            B = B ^ (1 << i);
        }
    }

    // Print A and B
    document.write(A + " " + B);
}

// Driver Code
var A = 7, B = 4;

flipBitsOfAandB(A, B);

// This code is contributed by Princi Singh

</script>

Output:

3 0

Time Complexity: O(32) 
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the following observations:

Find the bits which are set bits in both A and B using (A & B). 
Clear the bits of A which are set bits in both A and B using A = (A ^ (A & B)) 
Clear the bits of B which are set bits in both A and B using B = (B ^ (A & B)) 

Follow the steps below to solve the problem:

• Update A = (A ^ (A & B)).
• Update B = (B ^ (A & B)).
• Finally, print the updated values of A and B.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to flip bits of A and B
// which are set in both of them
void flipBitsOfAandB(int A, int B)
{

    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);

    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);

    // Print updated A and B
    cout << A << " " << B;
}

// Driver Code
int main()
{
    int A = 10, B = 20;

    flipBitsOfAandB(A, B);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to flip bits of A and B
// which are set in both of them
static void flipBitsOfAandB(int A, int B)
{

    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);

    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);

    // Print updated A and B
    System.out.print(A + " " + B);
}

// Driver Code
public static void main(String[] args)
{
    int A = 10, B = 20;

    flipBitsOfAandB(A, B);
}
}

// This code is contributed by sanjoy_62

Python3

# Python3 program to implement
# the above approach

# Function to flip bits of A and B
# which are set in both of them
def flipBitsOfAandB(A, B):

    # Clear the bits of A which
    # are set in both A and B
    A = A ^ (A & B)

    # Clear the bits of B which
    # are set in both A and B
    B = B ^ (A & B)

    # Print updated A and B
    print(A, B)

# Driver Code
if __name__ == "__main__" :

    A = 10
    B = 20

    flipBitsOfAandB(A, B)

# This code is contributed by Virusbuddah_

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to flip bits of A and B
// which are set in both of them
static void flipBitsOfAandB(int A, int B)
{

    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);

    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);

    // Print updated A and B
    Console.Write(A + " " + B);
}

// Driver Code
public static void Main(String[] args)
{
    int A = 10, B = 20;

    flipBitsOfAandB(A, B);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// javascript program to implement
// the above approach


// Function to flip bits of A and B
// which are set in both of them
function flipBitsOfAandB(A , B)
{

    // Clear the bits of A which
    // are set in both A and B
    A = A ^ (A & B);

    // Clear the bits of B which
    // are set in both A and B
    B = B ^ (A & B);

    // Print updated A and B
    document.write(A + " " + B);
}

// Driver Code
var A = 10, B = 20;

flipBitsOfAandB(A, B);

// This code contributed by shikhasingrajput

</script>

Output:

10 20

Time Complexity: O(1) 
Auxiliary Space: O(1)